x = -5 - 4t
y = 1 - t
z = 3 +2t
and x + 2y + 3z - 9 = 0
Now how do I approach a problem like this? I am capable of calculating dot products, cross products, finding vectors from two points.
What is involved in this? Just a little boost will help and I'll try to solve it. Thanks!
This is where I have trouble. I'm used to dealing with points, I have difficulty reading the way the information is displayed. I'm not really sure which one is the plane.
Okay, I believe x + 2y + 3z - 9 = 0 is the plane and the other info can be used to find a point. How is this read? I know it is a parametric equation of the line, so it should be
x = -5
y = 1
z = 3
If I am reading this correctly, I can use the distance between point and a plane formula and we get -3 / square root 14.
Now that we have the distance, I'm supposed determine if the line and plane are parallel. I could use some help on this part.
You need to extensively review your class notes. Below are things you are expected to know.
The cartesian equation of a plane can be written ax + by + cz = d and a normal vector to this plane is <a, b, c>.
The parametric equations of a line can be written , , and a vector in the direction of this line is .
Now apply the defintion you will undoubtedly have been given on establishing whether a line and plane are parallel.
Okay, I think I got this. My teacher and textbook use different notations, or there are different notations so the examples I have in class are not exactly as those in the textbook but I think I got it.
With
x = -5 - 4t
y = 1 - t
z = 3 +2t
and x + 2y + 3z - 9 = 0
I can sub the x,y,z into the equation. Then in order to find out if it is parallel I have this note:
at + b = 0
a = 0, then b = 0
If we get 0 = 0 then any t is a solution and is parallel.
If we get b = 0 or b =/= 0 then perpendicular.
In this problem, we get
(-5-4t) + 2(1-t) + 3(3+2t) - 9 = 0
-5 -4t + 2 - 2t + 9 + 6t - 9 = 0
Since t = 0 then a = 0 and so b is 0. So by the above notes we should get 0 = 0. I actually get -3=0.
Following my notes, I would say this is parallel. I just don't understand why I get -3 = 0.
I don't know either. My teacher speaks with a thick Russian accent and understanding the material is hard enough in itself. Please bear with me.
Anyways, I believe I finally got it. I'm learning more on this subject with google and this website than my actual class. Thank god for this forum.
So, again, we have
x = -5 - 4t
y = 1 - t
z = 3 +2t
and x + 2y + 3z - 9 = 0
The vector of the line is (-5,1,3) and it's direction vector is (-4,-1,2)
The vector of the plane is (1,2,3).
I take the dot product, if it is 0 then they are parallel.
The dot product is -4 -2 +6 = 0 and so here we get the 0 = 0.
I'm going to have to clarify the notes with my teacher, chances are I'm just confusing two different things.